Is the frequency phase-response of a linear filter always decreasing? Consider a linear electronic filter with a sinusoidal input. Is it always true that the graph of phase vs input frequency is decreasing?
Note that I'm talking about physical passive electronic filters not mathematical constructions. Why is it that all the filters I've seen which are composed of resistors, capacitors, and inductors, have decreasing phase?
 A: No. In general the phase of minimum-phase filters and mixed-phase transfer functions is not a monotonic function of frequency.
Factorize a general rational transfer function into a ratio of the product of single pole and single zero transfer functions. Witness that the phase of each factor is additive (being the imaginary part of the logarithm).
Now sketch the phase (argument) of a vector joining a point on the imaginary axis ($s=i\,\omega$) as $\omega$ moves from $-\infty$ to $+\infty$.
You should readily see that:


*

*A pole in the left half plane yields a factor whose phase monotonically decreases with frequency;

*A zero in the left half plane yields a factor whose phase monotonically increases with frequency;

*A zero in the right half plane yields a factor whose  phase monotonically decreases with frequency;

*A pole in the right half plane yields a factor whose phase monotonically increases with frequency (although this is seldom useful, leading to an unstable transfer function).


In general, only maximum phase or all pole transfer functions have phase that monotonically decreases with frequency.
A: Here is one counter example - hope it works !
Consider a circuit made of two resistors $R$, two coils $L$ and one capacitor $C$ such that $LC=\omega_0^2=1$ and $\frac{1}{R}\sqrt{\frac{L}{C}}=Q=1$, all connected in series. The circuit is powered by $V_{in}=V_0 e^{i\omega t}$ and the out voltage is taken over one coil, one capacitor and the resistor (see scheme).

The transfer function is given by
$$
\frac{V_{out}}{V_0} = \frac{Z_{R+L+C}}{Z_{R+L+C}+Z_{L+R}}=\frac{\frac{1}{jC\omega}+R+jL\omega}{\frac{1}{jC\omega}+2R+2jL\omega}=\frac{1+ix-x^2}{1+2ix-2x^2}
$$
and the phase is first decreasing, then increasing as shown below

